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(i) Prove that : $\sqrt{\sec^2\theta + \operatorname{cosec}^2\theta} = \tan\theta + \cot\theta$
(ii) Evaluate: $\frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$
(ii) Evaluate: $\frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$
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(i) LHS $= \sqrt{1 + \tan^2\theta + 1 + \cot^2\theta}$
$= \sqrt{\tan^2\theta + \cot^2\theta + 2 \times \tan\theta \times \cot\theta}$
$= \sqrt{(\tan\theta + \cot\theta)^2}$
$= \tan\theta + \cot\theta = \text{RHS}$
(ii) $\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$
$= \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1+\sqrt{3})}$
$= \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} \times \frac{\sqrt{2}(1-\sqrt{3})}{\sqrt{2}(1-\sqrt{3})}$
$= \frac{\sqrt{6}(1-\sqrt{3})}{4(1-3)} = \frac{\sqrt{6}-\sqrt{18}}{-8} = \frac{3\sqrt{2}-\sqrt{6}}{8}$
$= \sqrt{\tan^2\theta + \cot^2\theta + 2 \times \tan\theta \times \cot\theta}$
$= \sqrt{(\tan\theta + \cot\theta)^2}$
$= \tan\theta + \cot\theta = \text{RHS}$
(ii) $\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$
$= \frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1+\sqrt{3})}$
$= \frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})} \times \frac{\sqrt{2}(1-\sqrt{3})}{\sqrt{2}(1-\sqrt{3})}$
$= \frac{\sqrt{6}(1-\sqrt{3})}{4(1-3)} = \frac{\sqrt{6}-\sqrt{18}}{-8} = \frac{3\sqrt{2}-\sqrt{6}}{8}$